1 /** 2 * Definition for binary tree with next pointer. 3 * struct TreeLinkNode { 4 * int val; 5 * TreeLinkNode *left, *right, *next; 6 * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 void connect(TreeLinkNode *root) { 12 TreeLinkNode* cur = root; 13 while(cur) { 14 TreeLinkNode* node = cur; 15 TreeLinkNode* last = NULL; 16 cur = NULL; 17 while(node) { 18 TreeLinkNode* left = node->left; 19 TreeLinkNode* right = node->right; 20 if(left || right) { 21 if(last) last->next = left ? left : right; 22 if(left) left->next = right; 23 if(!cur) cur = left ? left : right; 24 last = right ? right : left; 25 } 26 node = node->next; 27 } 28 } 29 } 30 };
Follow up for problem "Populating Next Right Pointers in Each Node".
What
if the given tree could be any binary tree? Would your previous solution still
work?
Note:
You may only use constant extra space.
For example,
Given the following binary tree,
1
/ \
2 3
/ \
\
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4->
5 -> 7 -> NULL
Solution: 1. iterative way with CONSTANT extra space.
2.
iterative way + queue. Contributed by SUN Mian(孙冕).
3. tail recursive
solution.
*/
Populating Next Right Pointers in Each Node II,布布扣,bubuko.com
Populating Next Right Pointers in Each Node II
原文:http://www.cnblogs.com/zhengjiankang/p/3680276.html